/*
Advanced Fruits
Problem Description
The company "21st Century Fruits" has specialized in creating new sorts of fruits by transferring genes from one fruit into the genome of another one. Most times this method doesn't work, but sometimes, in very rare cases, a new fruit emerges that tastes like a mixture between both of them.
A big topic of discussion inside the company is "How should the new creations be called?" A mixture between an apple and a pear could be called an apple-pear, of course, but this doesn't sound very interesting. The boss finally decides to use the shortest string that contains both names of the original fruits as sub-strings as the new name. For instance, "applear" contains "apple" and "pear" (APPLEar and apPlEAR), and there is no shorter string that has the same property.
A combination of a cranberry and a boysenberry would therefore be called a "boysecranberry" or a "craboysenberry", for example.
Your job is to write a program that computes such a shortest name for a combination of two given fruits. Your algorithm should be efficient, otherwise it is unlikely that it will execute in the alloted time for long fruit names.
Input
Each line of the input contains two strings that represent the names of the fruits that should be combined. All names have a maximum length of 100 and only consist of alphabetic characters.
Input is terminated by end of file.
Output
For each test case, output the shortest name of the resulting fruit on one line. If more than one shortest name is possible, any one is acceptable.
Sample Input
apple peach
ananas banana
pear peach
Sample Output
appleach
bananas
peachr
题意：将两个字符串结合起来，他们的公共子串只输出一次
思路：根据LCS的原理，将每个字符都进行标记，看两个字符串中对应的字符究竟处于什么状态，然后输出，其标记为公共子串的字符只输出一次即可
*/
#include <bits/stdc++.h>
#define ll long long
#define PI acos(-1)
#define M(n, m) memset(n, m, sizeof(n));
const int INF = 1e9 + 7;
const int maxn = 1e3 + 10;
using namespace std;

int dp[maxn][maxn], mark[maxn][maxn];
string s1, s2;

// 通过mark数组初始化的不同来选择输出哪个字符串
void print(int x, int y)
{
    if (!x && !y)
        return;
    if (mark[x][y] == 0)
    {
        print(x - 1, y - 1);
        cout << s1[x - 1];
    }
    else if (mark[x][y] == 1)
    {
        print(x - 1, y);
        cout << s1[x - 1];
    }
    else
    {
        print(x, y - 1);
        cout << s2[y - 1];
    }
}

void LCS()
{
    // 边界初始化
    for (int i = 0; i <= s1.size(); i ++)
        mark[i][0] = 1;
    for (int i = 0; i <= s2.size(); i ++)
        mark[0][i] = -1;

    for (int i = 1; i <= s1.size(); i ++)
        for (int j = 1; j <= s2.size(); j ++)
        {
            if (s1[i - 1] == s2[j - 1])
            {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                mark[i][j] = 0;
            }
            else if (dp[i - 1][j] >= dp[i][j - 1])
            {
                dp[i][j] = dp[i - 1][j];
                mark[i][j] = 1;
            }
            else
            {
                dp[i][j] = dp[i][j - 1];
                mark[i][j] = -1;
            }
        }
}

int main()
{
    while (cin >> s1 >> s2)
    {
        M(dp, 0)
        LCS();
        print(s1.size(), s2.size());
        cout << endl;
    }
    return 0;
}
